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1
1
1
0.750
75.0%
4
a
c
T
T
ε
=− =− =
=
where the gas law in ratio form has been used.
(e) This is greater than our result in part (c), as expected from the second law of
thermodynamics.
31. (a) The net work done is the rectangular “area” enclosed in the
pV
diagram:
()
(
)
(
)
00
0
0
0
0
0
0
22
.
WV
Vp
p
V
V
pp
V
p
=−
−
=
−
−
=
Inserting the values stated in the problem, we obtain
W
= 2.27 kJ.
(b) We compute the energy added as heat during the “heatintake” portions of the cycle
using Eq. 1939, Eq. 1943, and Eq. 1946:
()()
35
+1
+
3
5
1+
2 1+ 4 2
2
2
13
2
bc
b
abc
V
b
a
p
c
b
a
a
aa
a
b
a
a
TT
T
Qn
C
T
T
n
C
T
T
n
R
T
n
R
T
T
T
nRT
p V
T
pV
§·
§
·
−
=
−
−
¨¸
¨
¸
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Energy, Work, Heat

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